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# How do I use DeMoivre's theorem to solve ##z^3-1=0##?

If ##z^3-1=0##, then we are looking for the cubic roots of unity, i.e. the numbers such that ##z^3=1##.

If you're using complex numbers, then every polynomial equation of degree ##k## yields exactly ##k## solution. So, we're expecting to find three cubic roots.

De Moivre's theorem uses the fact that we can write any complex number as ##\rho e^{i \theta}= \rho (\cos(\theta)+i\sin(\theta))##, and it states that, if ##z=\rho (\cos(\theta)+i\sin(\theta))##, then ##z^n = \rho^n (\cos(n \theta)+i\sin(n \theta))##

If you look at ##1## as a complex number, then you have ##\rho=1##, and ##\theta=2\pi##. We are thus looking for three numbers such that ##\rho^3=1##, and ##3\theta=2\pi##.

Since ##\rho## is a real number, the only solution to ##\rho^3=1## is ##\rho=1##. On the other hand, using the periodicity of the angles, we have that the three solutions for ##\theta## are ##\theta_{1,2,3}=\frac{2k\pi}{3}##, for ##k=0,1,2##.

This means that the three solutions are:

- ##\rho=1, \theta=0##, which is the real number ##1##.
- ##\rho=1, \theta=\frac{2\pi}{3}##, which is the complex number ##-1/2 + \sqrt{3}/2 i##
- ##\rho=1, \theta=\frac{4\pi}{3}##, which is the complex number ##-1/2 - \sqrt{3}/2 i##